# With 90% confidence for the sample mean 391.00, the sample standard deviation 13.60, and sample...

## Question:

With {eq}90 \% {/eq} confidence for the sample mean {eq}391.00 {/eq}, the sample standard deviation {eq}13.60 {/eq}, and sample size {eq}35 {/eq}, what is the upper confidence limit?

## Confidence Interval:

The confidence interval can be obtained to determine the chances that the real parameter lies in the estimated limits at the pre-defined level of confidence. If it satisfies the randomization condition, then confidence interval can be constructed for normally distributed populations.

Given Information:

• The confidence level (CL) is 0.90.
• The sample mean {eq}\left( {\bar x} \right) {/eq} is 391.00.
• The sample standard deviation (s) is 13.60.
• The sample size (n) is 35.

Here, we use z distribution as the sample size (n) is more than 30.

The alpha value or level of significance {eq}\left( \alpha \right) {/eq} is given by:

{eq}\begin{align*} \alpha &= 1 - CL\\ &= 1 - 0.90\\ &= 0.10 \end{align*} {/eq}

The two tailed critical value {eq}\left( {{z^ * }} \right) {/eq} obtained from the z- table at the significance level 0.10 is +/- 1.645.

The upper confidence limit for the population mean can be calculated as:

{eq}\begin{align*} UCL &= \bar x + \left( {{z^ * } \times \dfrac{s}{{\sqrt n }}} \right)\\ & = 391.00 + \left( {1.645 \times \dfrac{{13.60}}{{\sqrt {35} }}} \right)\\ & = 391.00 + 3.78\\ & = 394.78 \end{align*} {/eq}

Therefore, the upper confidence limit for the population mean is 394.78.